Chapter 21, Problem 79A

To determine

Capacitor’s capacitance.

Answer to Problem 79A

  $C=2\mu F$

Explanation of Solution

Given:

Charge on capacitor, $C=90\mu \text{C}$

Electric potential difference, $\Delta V=45\text{V}$

Formula Used:

The capacitance of capacitor is, $C=\frac{q}{\Delta V}$

Where, $q$ is charge on the capacitor and $\Delta V$ is electric potential difference across capacitor.

Calculation:

On substituting the given values,

  $\begin{array}{l}\text{C}=\frac{90\times {10}^{-6}}{45}\\ =2\times {10}^{-6}F\end{array}$

Conclusion:

Therefore, the capacitor’s capacitance is $C=2\mu F$ .